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Math Help - continuous random variables

  1. #1
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    continuous random variables

    how would I compute the answers to the following questions:

    A supermarket manager finds that 20% of his customers buy brand X cornflakes. The other 80% buy brand Z.

    a) Ina ramdom sample of 10, what is the probability that at least 4 customers buy brand X? [ANS:0.121]

    b) In a random sample of 100 customers, what is the probability that between 15 and 30 customers (inclusively) buy brand X? [ANS:0.9119]

    c) What is the minimum number of boxes of brand X cornflakes that the manager should have in stock to ensure, with a probability of at least 0.95, that he can meet the brand X demands of a random sample of 225 customers? (Assume 1 box per customer) [ANS:55]
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  2. #2
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    Quote Originally Posted by skhan
    how would I compute the answers to the following questions:

    A supermarket manager finds that 20% of his customers buy brand X cornflakes. The other 80% buy brand Z.

    a) Ina ramdom sample of 10, what is the probability that at least 4 customers buy brand X? [ANS:0.121]
    Here we are dealing with the binomial distribution, where the probability of
    R successes in N trials is:

    <br />
P(R,N)={N \choose R}p^R (1-p)^{N-R}<br />
,

    where p is the probability of a success on a single trial, and:

    {N \choose R}=\frac{N!}{R!(N-R)!}

    is a binomial coefficient.

    Now the probability Prb(4+) that at least 4 customers out of 10 buy brand X is
    1 minus the probability that 3 or less customers buy brand X. So:

    <br />
Prb(4+)=1-P(0,10)-P(1,10)-P(2,10)-P(3,10)<br />
    <br />
=1-{10 \choose 0}0.2^00.8^10-{10 \choose 1}0.2^1 0.8^9 -{10 \choose 2}0.2^2 0.8^8-{10 \choose 3}0.2^3 0.8^7<br />

    <br />
=1-0.1074-0.2684-0.3020-0.2013=0.1209<br />

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by skhan
    how would I compute the answers to the following questions:

    A supermarket manager finds that 20% of his customers buy brand X cornflakes. The other 80% buy brand Z.

    b) In a random sample of 100 customers, what is the probability that between 15 and 30 customers (inclusively) buy brand X? [ANS:0.9119]
    This still a problem about the binomial distribution, nut now we are in
    large sample territory and we will use the normal approximation to the
    binomial distribution.

    In a sample of 100 the mean number who buy brand X is \mu=0.2\times 100=20, and the standard deviation \sigma=\sqrt{100\times 0.2 \times 0.8}=4.

    Now in the normal approximation to the binaomial distribution we appproximate:

    <br />
P_{binomial(p,N)}(R)\approx P_{normal(\mu,\sigma)}(R-0.5\le x < R+0.5)<br />

    Now here we are asked for the probability that between 15 and 30 cusyomers
    buy brand X, this probability is:

    <br />
P_{binomial(p,N)}(15\mbox{ to }30)\approx P_{normal(\mu,\sigma)}(15-0.5\le x < 30+0.5)<br />

    <br />
P_{normal(\mu,\sigma)}(15-0.5\le x < 30+0.5) =P_{normal(0,1)}\left(\frac{14.5-20}{4}\le x < \frac{30.5-20}{4}\right)
    =P_{normal(0,1)}(-1.375\le x<2.625)\approx 0.9111<br />

    RonL
    Last edited by CaptainBlack; May 21st 2006 at 03:55 AM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by skhan
    how would I compute the answers to the following questions:

    A supermarket manager finds that 20% of his customers buy brand X cornflakes. The other 80% buy brand Z.

    c) What is the minimum number of boxes of brand X cornflakes that the manager should have in stock to ensure, with a probability of at least 0.95, that he can meet the brand X demands of a random sample of 225 customers? (Assume 1 box per customer) [ANS:55]
    The critical z-score for a standard normal distribution, so that the probability
    of not-exceeding it is 0.95 is 1.645 .

    With a sample size of 225 costomers the mean number who buy brand X is
    0.2\times 225=45 , and the standard devistion is \sqrt{225 0.2 0.8}=6 .

    So the number of customers N corresponding to the critical z-score satisfies:

    \frac{N-45}{6}=1.645

    so N=54.89 or rounding up to an integer number of packs N=55 .

    RonL
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